MATH 221 Lecture 16

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 6 August 2012

Lecture 16

Example:$$ ${\displaystyle \text{Graph}f\left(x\right)=\frac{x^2-1}{x^3-4x}}$

Notes:
(a)	${\displaystyle y=\frac{x^2-1}{x^3-4x}=\frac{(x+1)(x-1)}{x(x^2-4)}=\frac{(x+1)(x-1)}{x(x+2)(x-2)}}$
(b)	If $x=1$ then $y=0$.
(c)	If $x=-1$ then $y=0$.
(d)	If $x=2^{+}$ then $y\to \infty$.	${\displaystyle \left(\frac{\text{pos}.\text{pos}}{\text{pos}.\text{pos}.\text{pos}}\right)}$
(e)	If $x=2^{-}$ then $y\to -\infty$.	${\displaystyle \left(\frac{\text{pos}.\text{pos}}{\text{pos}.\text{pos}.\text{neg}}\right)}$
(f)	If $x=0^{+}$ then $y\to \infty$.	${\displaystyle \left(\frac{\text{pos}.\text{neg}}{\text{pos}.\text{pos}.\text{neg}}\right)}$
(g)	If $x=0^{-}$ then $y\to -\infty$.	${\displaystyle \left(\frac{\text{pos}.\text{neg}}{\text{neg}.\text{pos}.\text{neg}}\right)}$
(h)	If $x=-2^{+}$ then $y\to \infty$.	${\displaystyle \left(\frac{\text{neg}.\text{neg}}{\text{neg}.\text{pos}.\text{neg}}\right)}$
(i)	If $x=-2^{-}$ then $y\to -\infty$.	${\displaystyle \left(\frac{\text{neg}.\text{neg}}{\text{neg}.\text{neg}.\text{neg}}\right)}$
(j)	${\displaystyle y=\frac{x^2-1}{x^3-4x}\text{is the same as}(-y)=\frac{{(-x)}^2-1}{{(-x)}^3-4(-x)}}$ So if we slip $y$ to $-y$ and $x$ to $-x$ the graph stays the same.
(k)	If $x\to \infty$ then $y\to 0^{+}$.
(l)	If $x\to -\infty$ then $y\to 0^{-}$.

Example:$$ $\text{Graph}\sqrt{x}+\sqrt{y}=1$.

Notes:
(a)	If we switch $x$ and $y$ this graph stays the same.
(b)	If $x=0$ then $\sqrt{y}=1,$ so $y=1^2=1$.
(c)	If we switch $y=0$ then $x=1$.
(d)	If $x=y$ then $\sqrt{x}+\sqrt{x}=1$ and $\sqrt{x}=\frac{1}{2},x=\frac{1}{4}$.
(e)	This graph should be similar to $x^2+y^2=1$ or $x+y=1$

Example:$$ ${\displaystyle \text{Graph}\frac{x^2-1}{x^2+1}=f\left(x\right)}$.

Notes:
(a)	${\displaystyle y=\frac{x^2-1}{x^2+1}=\frac{x^2+1-2}{x^2+1}=1-\frac{2}{x^2+1}}$

Example:$$ $\text{Graph}\ln (4-x^2)=f\left(x\right)$.

Notes:
(a)	$y=\ln (4-x^2)=\ln \left((2+x)(2-x)\right)=\ln (2+x)+\ln (2-x)$.
(b)	If we flip $x$ to $-x$ the graph stays the same since $y=\ln (4-x^2)=\ln (4-{(-x)}^2)$

$y=\ln (4-x^2)=\ln (2+x)+\ln (2-x)$.

Example:$$ $\text{Graph}y=x^{\frac{2}{3}}{(6-x)}^{\frac{1}{3}}$

Notes:
(a)	If $x=0$ then $y=0$.
(b)	If $x=6$ then $y=0$.
(c)	If $x\to \infty$ then $y\to x^{\frac{2}{3}}{(-x)}^{\frac{1}{3}}=-x$.
(b)	If $x\to -\infty$ then $y\to \infty (y\to -x\text{again})$.

Example:$$ $\text{Graph}y=f\left(x\right)\text{when}x=\cos 2\theta \text{and}y=\cos \theta$.

Notes:

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta ={\cos }^2\theta -(1-{\cos }^2\theta )=2{\cos }^2\theta -1$

$\text{So this problem is really asking for the graph of}\cos 2\theta ={\cos }^{2}2\theta -1\text{when}x=\cos 2\theta \text{and}y=\cos \theta$
$\text{i.e.}x=2y^2=1$

${\displaystyle \text{If}x=2y^2-1\text{.}\text{Then}x+1=2y^2\text{.}\text{So}{y}^2=\frac{x+1}{2}\text{.}\text{So}y=\sqrt{\frac{x+1}{2}}\text{.}\text{So}f\left(x\right)=\sqrt{\frac{x+1}{2}}}$

Notes and References

These are a typed copy of lecture notes given by Arun Ram on October 13, 2000.

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